Completing the Square | 配方法

📚 Completing the Square | 配方法

Completing the square is one of the most powerful algebraic techniques in A-Level mathematics. It transforms a quadratic expression ax² + bx + c into the form a(x + h)² + k, revealing crucial information about the graph, the solutions, and the range of the function. Mastery of this technique is essential for success in Edexcel A-Level Mathematics, as it appears in topics ranging from solving equations to integration and even in mechanics problems involving projectile motion.

配方法是A-Level数学中最强大的代数技巧之一。它将二次表达式 ax² + bx + c 转化为 a(x + h)² + k 的形式，从而揭示函数图像、方程解以及函数值域的关键信息。掌握这一技巧对于成功应对Edexcel A-Level数学考试至关重要，它出现在从解方程到积分乃至涉及抛体运动的力学问题等各个专题中。

1. The Core Idea | 核心理念

The identity that forms the foundation of completing the square is (x + p)² = x² + 2px + p². If we have x² + bx, we can reverse this expansion. Notice that in the expansion, the coefficient of x is 2p, so p = b/2. Therefore, x² + bx = (x + b/2)² − (b/2)².

构成配方法基础的恒等式是 (x + p)² = x² + 2px + p²。如果我们有 x² + bx，可以逆向运用这一展开式。注意到在展开式中，x 的系数是 2p，因此 p = b/2。所以，x² + bx = (x + b/2)² − (b/2)²。

x² + bx = (x + b/2)² − b²/4

The key insight is that we “complete” the square by adding (b/2)², but to keep the expression unchanged, we must also subtract it. This technique works for any quadratic expression where the coefficient of x² is 1. Visually, we are rewriting the quadratic so that the variable x appears exactly once, making the expression easier to analyse and manipulate.

关键的洞察在于，我们通过加上 (b/2)² 来”配成”完全平方，但为了保持表达式不变，必须同时减去它。该技巧适用于任何 x² 系数为 1 的二次表达式。从视觉上看，我们重写了二次式，使变量 x 恰好出现一次，从而使表达式更易于分析和操作。

2. Worked Example 1: Monic Quadratic | 示例1：首项系数为1的二次式

Consider the expression x² + 6x + 5. Here, b = 6, so b/2 = 3. We add and subtract 3² = 9:

考虑表达式 x² + 6x + 5。这里 b = 6，所以 b/2 = 3。我们加上并减去 3² = 9：

x² + 6x + 5 = (x² + 6x + 9) − 9 + 5 = (x + 3)² − 4

We can verify this by expanding: (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5. ✓ The completed square form immediately tells us that the vertex of the graph y = x² + 6x + 5 is at (−3, −4). The minimum value of the expression is −4, achieved when x = −3.

我们可以通过展开验证： (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5。✓ 配平方形式立即可告诉我们图像 y = x² + 6x + 5 的顶点在 (−3, −4)。该表达式的最小值为 −4，在 x = −3 时取得。

Notice the pattern: for a monic quadratic x² + bx + c, we always use the formula x² + bx + c = (x + b/2)² − (b/2)² + c. This single formula handles every monic case quickly and reliably.

注意模式：对于首项系数为1的二次式 x² + bx + c，我们始终使用公式 x² + bx + c = (x + b/2)² − (b/2)² + c。这个公式可以快速可靠地处理所有此类情形。

3. Worked Example 2: Non-Monic Quadratic | 示例2：首项系数不为1的二次式

When the coefficient of x² is not 1, we first factor it out from the x² and x terms, then complete the square for the remaining expression. The factor must only be applied to the first two terms; the constant term is left outside until the final simplification.

当 x² 的系数不为 1 时，我们首先将其从 x² 项和 x 项中提取出来，然后对剩余的表达式配方。该因子只作用于前两项；常数项在最终化简前保持在外。

Consider 2x² − 8x + 7. First, factor out 2 from the first two terms:

考虑 2x² − 8x + 7。首先从前两项中提取因子2：

2x² − 8x + 7 = 2(x² − 4x) + 7

Now complete the square inside the brackets: x² − 4x = (x − 2)² − 4. Substituting back:

现在对括号内的部分配方：x² − 4x = (x − 2)² − 4。代回原式：

2x² − 8x + 7 = 2[(x − 2)² − 4] + 7 = 2(x − 2)² − 8 + 7 = 2(x − 2)² − 1

The vertex is at (2, −1) and the minimum value is −1. Note how the factor 2 multiplies the bracket, so it also multiplies the −4 term. This is a common source of errors, so always expand your final answer to check.

顶点在 (2, −1)，最小值为 −1。注意因子2乘以括号内的每一项，因此也乘以 −4 项。这是常见的错误来源，因此务必展开最终答案进行验证。

4. The General Formula | 一般公式

For any quadratic ax² + bx + c where a > 0, the completed square form is:

对于任意 a > 0 的二次式 ax² + bx + c，配平方形式为：

ax² + bx + c = a(x + b/2a)² + c − b²/4a

This is the general formula, though in practice it is usually easier to work through the steps rather than memorise this expression. The vertex of the parabola is at (−b/2a, c − b²/4a). The derivation follows directly from factoring out

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